MATHEMATICS
PAPER 1
Instructions to candidates
 This paper contains TWO sections: section I and section II
 Answer ALL the questions in Section I and only five questions from section II.
 Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
 Marks may be given for correct working even if the answer is wrong.
 Nonprogrammable silent electronic calculators and KNEC mathematical tables may be used except where stated otherwise.
Questions
SECTION I (50Marks) Answer ALL questions in this section
 Evaluate without using tables or a calculator the value of 1.33×0.510.19×0.0017 (3marks)
 When a certain number is divided by 48, 72 or 100 the remainder is 3 in each case. Find the number. (3mks)
 Find all the integral values of x which satisfy the inequalities (3mks)
20x>5+2x≥x+5  A Kenyan bank buys and sells foreign currency as shown below.
Buying Selling
Kenya shillings Kenya shillings
1 Euro 84.15 84.26
1 US Dollar 80.12 80.43
A tourist travelling from Britain arrives in Kenya with 5000 Euros. He converts all the Euros to Kenya Shillings at the bank. While in Kenya he spends a total of KSh. 289,850 and then converts the remaining Kenya shillings to US dollars at the bank. Calculate (to nearest dollar) the amount he receives? (3mks)  Complete the figure below so as to make the net of a cuboid. Hence determine the surface area of the cuboid. (4 Marks)
 Find the value of x given that is a singular matrix (3 marks)
 The figure below shows a solid wedge PQRSTU. Complete the solid showing all the hidden edges with dotted lines. (3marks)
 During an annual general meeting at Patel Mixed Day, goats and chicken were slaughtered. The number of heads for both chicken and goats were 45. The total number of legs were 100. Determine the exact number of goats and chicken slaughtered. (3marks)
 In a mixed school there are 900 students, out of these 600 are girls.
 Find the ratio of boys to girls. (2marks)
 What is the percentage of boys in this school? (1mark)
 Find the value of t in the equation (3marks)
t1  4+t =0
3 4  Using tables, find the reciprocal of 0.432 and hence evaluate (3marks)
√0.1225
0.432  Find the equation of the line perpendicular to 3x – 7y – 20 = 0, and passes through the point (5, 2). (3 Marks)
 The angle of elevation of the top of a building from a point P is 45°. From another point T, 15 meters nearer the foot of the building, the angle of elevation of the top of the building is 52°. Calculate the height of the building. (3marks)
 Given A = (10 7 5 9 11 12) and B=( 3 4 1 6 0 3 ), find AB (4mks)
 The travel timetable below shows the departure and arrival time for a bus plying between two towns M and R, 300 kilometres apart.
TOWN ARRIVAL DEPARTURE
M 0830h
N 1000h 1020h
P 1310h 1340h
Q 1510h 1520h
R 1600h
Calculate the average speed for the whole journey. (3mks)  Evaluate the value of x in 81^{x+1}+3^{4x}=243 (3marks)
SECTION II (50 Marks) Answer only FIVE questions from this section.
 A group of choir members decided to raise 3600/= to buy a guitar. Each member was to contribute equal amount. In the preparation process five members transferred to another church which meant the remaining contributors had to pay more to achieve the target.
 Show that the increase in the contribution per member was:
Sh. 18,000
n(n5)
if n is the initial number of members. (4 mks)  If the increase in the contribution per member was sh. 24, what was the original contribution before the other members left? ( 4 mks)
 Calculate the percentage increase in the contribution after the others left. (2 mks )
 Show that the increase in the contribution per member was:
 The table below shows the age groups and number of people who are HIV/ AIDS positive in a certain Subcounty in Kenya.
Age group 1019 2029 3039 4049 5059 6069 7079 No. of people 12 15 16 25 18 10 4  State the modal age group. (1mark)
 Calculate the mean age of the people who are HIV/AIDS Positive. (3marks)
 Calculate the median of the age group. (3marks)
 Using a pair of compass and ruler only construct.
 Triangle PQR in which PQ =5cm, QPR = 300 and PQR=105°. (3marks)
 A circle that passes through the vertices of the triangle PQR. Measure its radius. (3marks)
 The height of triangles PQR with PQ as the base. Measure the height. (2marks)
 Determine the area of the circle that has outsides the triangle correct to 2 decimal places (2marks)
 The diagram below shows a solid made of a hemisphere and a cylinder. The radius of both the cylinder and the hemisphere is 3cm. The length of the cylinder is 12cm.

 Calculate the volume of the solid. (3marks)
 The solid fits in a box in the shape of a cuboid 15 cm by 6cmby 6cm. Calculate the volume of the box not occupied by the solid correct to four significant figures. (2marks)

 Calculate the total surface area of the solid correct to four significant figures. (3marks)
 The surface of the solid is to be painted. One millilitre of paint covers an area of 8cm2.The cost of paint is Ksh 900 per litre. Calculate the cost of the paint required. (2marks)

 Find
 The surface area of the frustrum (5 Marks)
 The volume of frustrum shown. (5 Marks)
 A pirate boat sails from port A on a bearing of 050° at a speed of 112km/h, for 2½ hours to port B. From port B it changes its course and travelled on a bearing of 170° at a speed of 75km/h for 2^{2}/_{3} hours toward part C. From C it travelled to port D. D is on a bearing of 130° and 160km from A.
 Using a scale of 1cm to represent 40km, Draw a diagram showing the positions of the ports A, B, C and D. (4 marks)
 Use your drawing to find.
 The distance CD (1 mark)
 The bearing of C from D (1 mark)
 A marine police patrol leaves port A to intercept the pirate boat at M as it moves from B to C in the shortest time possible.
 How far from A will the two boats meet at M? (2 marks)
 If the boats meets after 2 hours, what is the speed of the marine police patrol boat? (2 marks)
 Helena left town A at 8:00 am and travelled towards town B at an average speed of 64 km/h. half an hour later, Joan left town B and travelled towards A at the same speed if the two towns are 384km apart:
 At what time of the day did they meet? (5marks)
 How far from town B was their meeting point? (2marks)
 How far apart were they at 10:30am (3marks)
 The figure below shows two intersecting circles with centres A and C, of radius 5cm and 8cm respectively. The common chord EF = 6cm
Calculate Angle EAF (2 marks)
 Angle ECF (2 marks)
 Calculate the common area between the two intersecting circles. (6 marks)
Marking Scheme
 133 X 0.51 X 1000000
0.19 X 0.0017 X 1000000
133 X 51 X 100
19 X 17
2100 
2 48 72 100 2 24 36 50 2 12 18 25 2 6 9 25 3 3 9 25 3 1 3 25 5 1 1 25 5 1 1 5 1 1 1
= 3600
Number =3600 + 3
= 3603  20 – x > 5+2x
15 >3x
x <5
5 + 2x ≥ x + 5
2x – x ≥ 0
3x ≥ 0
x ≥0
5> x ≥ 0
Integral values are 0, 1, 2, 3 and 4  5000 x 84.15 = 420,750
420 750
289 850
130 900
= 1627.50
≈ 1628  Surface area =
3 x 5 x 2 = 30
2 x 5 x 2 = 20
2 x 3 x 2 = 12
= 62cm^{2}  2x  1  x² = 0
x²  2x + 1 = 0
(x  1) (x1) = 0
x = 1  Diagram
 G+c=45
4g+2c= 100
G=45c
4(45 – c) + 2c=100
180 – 4c +2c=100
C=40
g=5 
 boys = 900 – 600= 300
ratio
3000:600
1:2  300/900 x 100
33^{1}/_{3}% M1
 boys = 900 – 600= 300
 4(t1) – 3(4+t)=0
4t3t – 4 – 12=0
t 16 = 0
t= 16  1 = 1x 1
0.432 4.32 x 10^{1}
= 0.2375 x 101
= 2.315
= √0.1225 = 2.315x√0.12225
0.432
= 0.35 x 2.315
= 0.81025  7y = 3x – 20
y = x 
g =
Gradient of tar = ∆y/∆x =
= 3y – 6 = 7x + 35
3y = 7x + 41
y = x +  diagram
x = h/tan 52
x = h
tan45 15
h = h
tan52 tan 4515
h(10.7813) = 15 h =
h = 68.58m 
 Time taken 1600 h – 830 h = 7hrs 30 min
= 7^{1}/_{2} hrs
Av. speed = 300/7.5
= 40km/h  3^{4(x+1)} +3^{4x} = 246
3^{4x+4} +3^{4x} = 246
3^{4x}(3^{4}+1)=246
3^{4x}=246
81
3^{4x}=3^{1}
4x= 1
X=^{1}/_{4} 
 Original contribution = ^{3600}/_{n}
New contribution = ^{3600}/_{n5}
Increase = ^{3600}/_{n5} = ^{3600}/_{n}
3600n + 3600n + 18000
n(n5)  18000 = 24
n(n5)
18000 = 24n^{2} 120n
n^{2} – 5n – 750 = 0
n^{2} – 30n + 25n – 750 = 0
n (n – 30) + (25 (n – 30 ) = 0
(n – 30) (n + 25) = 0
n  30 = 0
n + 25 = 0
but n cannot be –ve n = 30  3600/n= 120 original
3600/25= 144 new
increase = ^{24}/_{100} x 100% = 20 % (10 mks)
 Original contribution = ^{3600}/_{n}

 40 49

x f fx cf 14.5
24.5
34.5
44.5
54.5
64.5
74.512
15
16
25
18
10
4174
3675
552
1112.5
981
645
29812
27
43
68
86
96
100Σf=100 Σfx= 4130
x= Σfx/Σf = 4130/100 = 41.30  Median = 39.5 + ( ¹⁰⁰/₂  43) x 10
25
39.5 + 2.8


 Radius = 3.5 ± 0.1
 height construction
height = 3.4±0.1
 area of circle outside triangle
= ^{22}/_{7} x 3.5 – ½ x 3.4 x 5
= 29.98



 v=3.142 x 3^{2} x 12+12+^{2}/_{3} x 3.142 x 3^{3}
=339. 336 + 56.556
= 395.892
= 395.9  v= 15 x 6 x6 – 395.892
= 144.108
=144.1
 v=3.142 x 3^{2} x 12+12+^{2}/_{3} x 3.142 x 3^{3}

 S.A = 3.142 x 3^{2} +2 x3.142 x 3 x 12 + 2 x 3.14 2 x 3^{2}
= 28 .278 +226.224+56.556
311.058
= 311.1  Cost = 311.058 x 900
8x1000
= ksh. 34.99
= 35.0
 S.A = 3.142 x 3^{2} +2 x3.142 x 3 x 12 + 2 x 3.14 2 x 3^{2}


 Linear scale factor (L.S.F)
^{2.8}/_{3.5} = ^{4}/_{5}
Area scale factor (A.S.F) ( )2 =
Volume scale factor (V.S.F)
(^{4}/_{5})^{3} = ^{64}/_{125}
From similar triangles
^{h}/_{h+5} = ^{4}/_{5}
5h = 5h + 20
h = 20cm
Length of larger cone
L^{2} = 25^{2} + 3.5^{2} = 625 + 12.25
=637.25
L = √637.25
∴ l = 25.24
Curved surface area larger cone
πrL=^{22}/_{7} x 3.5 x 25.24
= 277.64cm^{2}Curved S.A of the small cone
^{16}/_{25} x 277.64 = 99.9cm2
Total surface area of frustrum
{22/7 x 2.8 x 2.8} + {22/7 x 3.5 x 3.5} + 99.95 cm^{2}24.64 + 38.5 + 99.95
= 163.09cm^{2}  Volume of small cone
1/3r2h = 1/3 x 22/7 x 2.8 x 2.8 x 20 = 164.3cm^{3}
Using volume scale factor (V.S.F)
Volume of larger cone
= ^{125}/_{64} x 164.3cm^{3}
∴Volume of frustrum
^{61}/_{64}= x 164.3
= 156.6cm^{3}
 Linear scale factor (L.S.F)

 Dist. AB = 112 × 5/2 = 280km
BC = 75 × 8/3 = 200km
AB = 7cm, BC=5cm, AD=4cm 
 CD = 3.5 × 40 = 140KM
 Bearing of C from D is 075°

 AM = 6 × 40 = 240km
 Speed = 240/2
= 120km/h
 Dist. AB = 112 × 5/2 = 280km

 64 x 0.5 = 32 km
384 32 = 352 km
Time = Distance
Relative speed
= 352
64 x 2
= 2.75 hrs
= 2 hrs 45 min
meeting time = 8.30 + 2.45
= 1075
= 1115 h or 11:15am  D = S x T
= 64 x 2.75
= 176km  At 10:30; time difference = 10.30 – 8.30 = 2 hrs
64 x 2 x 64 x 2 = 352
x = 352 – 256 = 96 km apart
 64 x 0.5 = 32 km

Sin 1/2x = 3/5
1/2x = Sin1 3/5 = 36.87
<EAT = 73.74°
Sin ½θ = 3/8
½θ = Sin1m(3/8)
<= 22.02
<ECF = 44.05 °
73.74 x 22 x 5^{2}  ^{1}/_{2} x 5 x 5Sin 73.7
360 7
=160.09  12
4.09cm
44.05 x 22 x 8^{2}  ^{1}/_{2} x 64Sin44.05
360 7
24.61  22.25
2.36
Total area= 4.09 + 2.36
= 6.45cm
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